Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

I'd use the symmetry of the curves (i.e., deal with everything above the initial line, then double at the end).

Work out the area bounded by the sector above the initial line (C2 formula), and then integrate the curve C1 between pi/3 and pi.

... and then double your answer to that.

Hi, does anyone know please whether the integral in question 6 here is on the spec http://madasmaths.com/archive/maths_booklets/further_topics/integration/1st_order_differential_equations_substitutions.pdf as I've not seen an integral that leads to arctan as the answer before? Thanks

Please could anyone explain question 11, especially part c, really struggling with these complex types! Thank you @Zacken

From part a(i) and a(ii) you get lambad =3 and mu=-sqrt(3)
From the sketch you did in part b, you have 4 points that belong on the circle.
So, you can subistute the cartesian equivlents of these points into the equation for a circle (x-a)^2 + (y-b)^2 = r^2 and then solve silmutaneously. Because the 4 points lie on either the x or y axis, this makes it a lot easier to solve.
you should get centre (-0.5(1+sqrt(3)), 0.5(1+sqrt(3)))

Can anyone help me with part c please as it doesnt mention z so i dont know how to bring z-3i in

Thanks for your help how did you get the values for lambda and mu in the first part please? Thanks

sorry for the late reply
for part a(i), you can use argz= arctan(b/a) where b=complex part and a=real part.
since the number lamba has no real part, subsitute z=bi
split the arg(bi-i/bi+1) into arg(bi-i)-arg(bi+1) = pi/6
arg(bi-i)=arg(i(b+1))=pi/2 so now
pi/2-pi/6=pi/3=arg(bi+1)
tan both sides as arg(bi+1) = arctan (b)
therefore b=tanpi/3 = sqrt(3)

for part a(ii) the logic is the same but substute z=a where a=real part as mu lies on the real axis so imaginary part =0 and then split the arg again.
I can go through this if you want but I'll let you try part a(ii) first. Its quite an intuitive solution but Im pretty sure its correct.

June 2014 FP2 question 6 (b)

For question 6 (b) in the second line of working where does the -3 come from in the first bracket?

Because we're told in the question that the equation of the circle is in the form (u-3) squared +v squared= k squared Hope that helps

ohhhhhhh yes that helps, I get it now. Thank you very much

No problem, thank you I understand now up to splitting the args but how do you know the value of arg(bi-i) and arg(bi+1) I tried it for part ii aswell but got stuck at finding the values of the args again? Thanks for your help

Can anyone explain part c to me please? The mark scheme says x=-1/4 a then WX = 2a +1/4 a but I dont understand how they have got this. Thanks

June 2014 FP2 question 6 (b)

For question 6 (b) in the second line of working where does the -3 come from in the first bracket?

The distance from O horiztontally to the left of the box is 1/4a

The distance from O horizontally to the right of the box is r = a(1+1) = 2a by subbing in theta = 0 into the polar eq.

Then asding these two distances gives you the total didtance from the left to the right of the box.